結果

問題 No.2236 Lights Out On Simple Graph
ユーザー au7777
提出日時 2023-04-18 15:24:31
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,246 ms / 4,000 ms
コード長 3,825 bytes
コンパイル時間 4,325 ms
コンパイル使用メモリ 246,996 KB
実行使用メモリ 47,484 KB
最終ジャッジ日時 2024-10-13 14:26:08
合計ジャッジ時間 35,992 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 57
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
using namespace std;
typedef pair<ll, ll> P;
using namespace atcoder;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
#define USE998244353
#ifdef USE998244353
const ll MOD = 998244353;
using mint = modint998244353;
#else
const ll MOD = 1000000007;
using mint = modint1000000007;
#endif
const int MAX = 2000005;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd(ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm(ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    // 
    ll m = 0;
    ll M = 3000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll keta(ll num, ll arity) {
    ll ret = 0;
    while (num) {
        num /= arity;
        ret++;
    }
    return ret;
}
ll ceil(ll n, ll m) {
    // n > 0, m > 0
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}
vector<ll> compress(vector<ll>& v) {
    // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] 
    vector<ll> u = v;
    sort(u.begin(), u.end());
    u.erase(unique(u.begin(),u.end()),u.end());
    map<ll, ll> mp;
    for (int i = 0; i < u.size(); i++) {
        mp[u[i]] = i;
    }
    for (int i = 0; i < v.size(); i++) {
        v[i] = mp[v[i]];
    }
    return v;
}
ll v_to_ll(vector<int>& v) {
    ll ret = 0;
    for (ll i = 0; i < v.size(); i++) {
        if (v[i]) {
            ret += ((ll)1 << i);
        }
    }
    return ret;
}
int main() {
    int n, m;
    cin >> n >> m;
    vector<int> a(m);
    vector<int> b(m);
    for (int i = 0; i < m; i++) {
        cin >> a[i] >> b[i];
        a[i]--;
        b[i]--;
    }
    vector<int> c(n);
    for (int i = 0; i < n; i++) {
        cin >> c[i];
    }
    unordered_map<ll, int> um;
    int fonum = m / 2;
    int lanum = m - m / 2;
    for (int i = 0; i < (1 << fonum); i++) {
        vector<int> reverse(n, 0);
        int cnt = 0;
        for (int j = 0; j < fonum; j++) {
            if ((i >> j) & 1) {
                cnt++;
                reverse[a[j]] = 1 - reverse[a[j]];
                reverse[b[j]] = 1 - reverse[b[j]];
            }
        }
        ll hash = v_to_ll(reverse);
        if (um.find(hash) == um.end()) {
            um.insert({hash, cnt});
        }
        else {
            // um.insert({hash, min(cnt, um[hash])});
            um[hash] = min(cnt, um[hash]);
        }
    }
    int ans = 100000;
    for (int i = 0; i < (1 << lanum); i++) {
        vector<int> reverse = c;
        int cnt = 0;
        for (int j = 0; j < lanum; j++) {
            if ((i >> j) & 1) {
                cnt++;
                reverse[a[fonum + j]] = 1 - reverse[a[fonum + j]];
                reverse[b[fonum + j]] = 1 - reverse[b[fonum + j]];
            }
        }
        ll hash = v_to_ll(reverse);
        if (um.find(hash) != um.end()) {
            ans = min(ans, cnt + um[hash]);
        }
    }
    if (ans == 100000) {
        cout << -1 << endl;
    }
    else {
        cout << ans << endl;
    }
    return 0;
}
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